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03-27-2012 10:53 AM

Hello all,

I have a problem with the computation of the IFFT (FFT ( x)) where x is a vector.

I made 2 tests:

test1 : 4096x1 vector , resulting IFFT (FFT (x)) is fine.

test2 : 4000000x1 vector , resulting IFFT (FFT( x)) after about 50000 records began to be ko.

my code:

%let NN=%sysevalf(4096);

proc iml;

use temp.H_primo_test;

read all var {x} into N;

z=J(&NN,**1**,**.**);

z = fft(N);

f=J(&NN,**1**,**.**);

f= ifft(z/(&NN));

quit**;**

it appears that the increase in the number of records creates problems for the Fast Fourier Transform.

it's correct?

thank you!

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Posted in reply to trevi83

03-27-2012 11:26 AM

I don't know what you are seeing or what your data are, but here is a program that correctly computes the FFT and IFFT of a signal with 4,000,000 elements. As far as I can tell, the FFT is computing correctly on this large vector:

**proc** **iml**;

NN = **4000000**;

t = T(do(-**10**,**10**, **20**/(NN-**1**)));

y = sin(t) - **2***sin(**2***t) + **3***sin(**3***t) + **0.1***rannor(t);

z = fft(y);

f= ifft(z/NN);

maxdiff = max(abs(y-f));

print maxdiff;

maxdiff |
---|

5.5052E-9 |

It's possible that your data are causing instabilities of some sort. Check to see if they are blowing up (spanning many orders of magnitude) or are widely oscilating or are otherwise degenerate.

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Posted in reply to Rick_SAS

03-27-2012 12:03 PM

Thank you Rick.

Your maxdiff is E-9 depends on the elements of the input vector.

My vector is composed of very small values (E-3, E-6...) so the differences are consistent.

However, you confirm that SAS creates differences by calculating the IFFT(FFT))?